convergence to exponential with order 1/n We know that  limit $\left(1+\dfrac{x}{n}\right)^n$ converges to $e^x$ but how can we prove that limit $\left(1+\dfrac{x}{n}+o(\frac{1}{n})\right)^n$ converges to $e^x$. 
 A: Suppose that $f(n)=o(1/n)$.
Choose any $\epsilon\gt0$. There is an $N$ so that if $n\ge N$, then $\left|f(n)\right|\le\epsilon/n$. Therefore,
$$
e^{x-\epsilon}=\lim_{n\to\infty}\left(1+\frac{x-\epsilon}n\right)^n\le\lim_{n\to\infty}\left(1+\frac xn+f(n)\right)^n\le\lim_{n\to\infty}\left(1+\frac{x+\epsilon}n\right)^n=e^{x+\epsilon}
$$
Since $\epsilon\gt0$ was arbitrary, we have
$$
\lim_{n\to\infty}\left(1+\frac xn+f(n)\right)^n=e^x
$$

As gniourf_gniourf mentions, we also need to choose $N$ big enough so that for $n\ge N$,
$$
1+\frac{x-\epsilon}{n}\ge0
$$
since $x\le y\implies x^n\le y^n$ when $x,y\ge0$.
A: For $\alpha_n \to 0$ as $n\to \infty$ we have
$$\lim_{n\to \infty}\left(1+\frac{x}{n}+\alpha_n\right)^n=e^{\lim_{n\to \infty} (x+n\alpha_n)}=e^x,$$ where in the last equality we have used that $\alpha_n=o(1/n),$ that is, $\lim_n n\alpha_n=0.$ 
Edit


*

*Justification of the identity used:


Assume $\alpha_n\to 0, \beta_n\to \infty$ and that $\lim_n\to \infty$ exits. Since $e=\lim_n (1+\alpha_n)^{1/\alpha_n}$ we have that
$$\displaystyle \lim_n (1+\alpha_n)^{\beta_n}=\lim_n (1+\alpha_n)^{ \frac{\alpha_n\beta_n}{\alpha_n}}=\left(\lim_n (1+\alpha_n)^{1/\alpha_n}\right)^{\lim_n \alpha_n\beta_n}=e^{\lim_n \alpha_n\beta_n}.$$


*

*$o()$-notation:


$\alpha_n=o(1/n)$ means, by definition, that $\lim_n \frac{\alpha_n}{1/n}=0.$ But $\lim_n \frac{\alpha_n}{1/n}=\lim_n n\alpha_n$ and so it exists and is $0.$
A: Fix $x\in\mathbb{R}$.
Let $n\in\mathbb{N}^*$. Then:
$$\left(1+\frac xn+o\left(\frac1n\right)\right)^n=\exp\left(n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\right).$$
Now, since
$$\dfrac xn+o\left(\dfrac1n\right)\underset{n\to+\infty}\longrightarrow0$$
and
$$o\left(\dfrac xn+o\left(\dfrac1n\right)\right)\underset{n\to+\infty}=o\left(\dfrac 1n\right),$$
we have
$$\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\underset{n\to+\infty}=\frac xn+o\left(\frac1n\right),$$
hence
$$n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)\underset{n\to+\infty}=x+o(1),$$
hence
$$\lim_{n\to+\infty}n\ln\left(1+\frac xn+o\left(\frac1n\right)\right)=x.$$
Now you can conclude by composing by the exponential function (which is continuous).
